Question

1. brodbelt 3200428 1 10 pts possible

consider the following reaction:
\\(\ce{cacn2 + 3h2o -> caco3 + 2nh3}\\)
105.0 g \\(\ce{cacn2}\\) and 78.0 g \\(\ce{h2o}\\) are reacted. assuming 100% efficiency, which reactant is in excess and how much is leftover? the molar mass of \\(\ce{cacn2}\\) is 80.11 g/mol. the molar mass of \\(\ce{caco3}\\) is 100.09 g/mol.
\\(\boldsymbol{\circ}\\) 1. \\(\ce{h2o}\\); 7.20 g left over
\\(\boldsymbol{\circ}\\) 2. \\(\ce{cacn2}\\); 70.8 g left over
\\(\boldsymbol{\circ}\\) 3. \\(\ce{cacn2}\\); 7.20 g left over
\\(\boldsymbol{\circ}\\) 4. \\(\ce{h2o}\\); 10.7 g left over
\\(\boldsymbol{\circ}\\) 5. \\(\ce{h2o}\\); 70.8 g left over
\\(\boldsymbol{\circ}\\) 6. \\(\ce{cacn2}\\); 10.7 g left over

Explanation:

Step 1: Calculate moles of \( \text{CaCN}_2 \)

Molar mass of \( \text{CaCN}_2 = 80.11 \, \text{g/mol} \). Moles \( = \frac{\text{mass}}{\text{molar mass}} \), so \( n_{\text{CaCN}_2} = \frac{105.0 \, \text{g}}{80.11 \, \text{g/mol}} \approx 1.311 \, \text{mol} \).

Step 2: Calculate moles of \( \text{H}_2\text{O} \)

Molar mass of \( \text{H}_2\text{O} = 18.02 \, \text{g/mol} \). Moles \( n_{\text{H}_2\text{O}} = \frac{78.0 \, \text{g}}{18.02 \, \text{g/mol}} \approx 4.328 \, \text{mol} \).

Step 3: Determine stoichiometric ratio

From reaction \( \text{CaCN}_2 + 3\text{H}_2\text{O} ightarrow \text{CaCO}_3 + 2\text{NH}_3 \), 1 mol \( \text{CaCN}_2 \) reacts with 3 mol \( \text{H}_2\text{O} \).

Step 4: Find required moles of \( \text{H}_2\text{O} \) for \( \text{CaCN}_2 \)

For \( 1.311 \, \text{mol} \, \text{CaCN}_2 \), required \( \text{H}_2\text{O} = 1.311 \times 3 = 3.933 \, \text{mol} \).

Step 5: Compare with available \( \text{H}_2\text{O} \)

Available \( \text{H}_2\text{O} = 4.328 \, \text{mol} \), which is more than required. So \( \text{H}_2\text{O} \) is in excess? Wait, no—wait, check again. Wait, maybe \( \text{CaCN}_2 \) is limiting? Wait, no, let's check moles of \( \text{CaCN}_2 \) required for \( \text{H}_2\text{O} \).

Step 6: Moles of \( \text{CaCN}_2 \) required for \( \text{H}_2\text{O} \)

For \( 4.328 \, \text{mol} \, \text{H}_2\text{O} \), required \( \text{CaCN}_2 = \frac{4.328}{3} \approx 1.443 \, \text{mol} \). But we have only \( 1.311 \, \text{mol} \, \text{CaCN}_2 \), so \( \text{CaCN}_2 \) is limiting? Wait, no—wait, initial moles: \( \text{CaCN}_2 = 1.311 \), \( \text{H}_2\text{O} = 4.328 \).

Stoichiometric ratio: \( \frac{n_{\text{CaCN}_2}}{1} = 1.311 \), \( \frac{n_{\text{H}_2\text{O}}}{3} \approx 1.443 \). Since \( 1.311 < 1.443 \), \( \text{CaCN}_2 \) is limiting. Wait, no—wait, the smaller value means limiting. So \( \text{CaCN}_2 \) is limiting, so \( \text{H}_2\text{O} \) is in excess? Wait, no, wait: when \( \text{CaCN}_2 \) is limiting, it will react completely, and \( \text{H}_2\text{O} \) will have leftover. Wait, but let's recalculate.

Wait, required \( \text{H}_2\text{O} \) for \( \text{CaCN}_2 \): \( 1.311 \times 3 = 3.933 \, \text{mol} \). Available \( \text{H}_2\text{O} = 4.328 \, \text{mol} \). Leftover \( \text{H}_2\text{O} \) moles \( = 4.328 - 3.933 = 0.395 \, \text{mol} \). Mass of leftover \( \text{H}_2\text{O} = 0.395 \times 18.02 \approx 7.12 \, \text{g} \), close to 7.20 g. Wait, maybe calculation differences. Wait, let's do precise:

Moles of \( \text{CaCN}_2 = 105.0 / 80.11 = 1.3107 \, \text{mol} \).

Required \( \text{H}_2\text{O} = 1.3107 \times 3 = 3.9321 \, \text{mol} \).

Available \( \text{H}_2\text{O} = 78.0 / 18.01528 = 4.3296 \, \text{mol} \) (using 18.01528 for H₂O).

Leftover \( \text{H}_2\text{O} \) moles = 4.3296 - 3.9321 = 0.3975 mol.

Mass = 0.3975 × 18.01528 ≈ 7.16 g, close to 7.20 g. So \( \text{H}_2\text{O} \) is in excess, leftover ~7.20 g. Wait, but the options: option 1 is \( \text{H}_2\text{O} \); 7.20 g left over.

Wait, but earlier mistake: when we check, \( \text{CaCN}_2 \) is limiting (since required \( \text{CaCN}_2 \) for \( \text{H}_2\text{O} \) is 1.443 mol, but we have 1.311 mol, so \( \text{CaCN}_2 \) is limiting, so \( \text{H}_2\text{O} \) is in excess. Wait, no—wait, the limiting reactant is the one that is completely consumed. So \( \text{CaCN}_2 \) is limiting (since we have less \( \text{CaCN}_2 \) than required for \( \text{H}_2\text{O} \)). So \( \text{H}_2\text{O} \) is in excess. Then leftover \( \text{H}_2\text{O} \) is 78.0 g - (mass of \( \text{H}_2\text{O} \) reacted). Mass of \( \text{H}_2\text{O} \) reacted = 3.9321 mol × 18.01528 g/mol ≈ 70.8 g. So leftover \( \text{H}_2\text{O} = 78.0 - 70.8 = 7.20 g \). Ah, that's it! So \( \text{H}_2\text{O} \) is in excess, 7.20 g left over.

Answer:

1. \( \text{H}_2\text{O} \); 7.20 g left over